Vidaara.orgClass 12 · Chemistry
CodeVID-C12-03-T3-01
Electrolysis, Batteries & Corrosion — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all steps in numericals and write balanced electrode reactions where asked. For full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
In an electrolytic cell, reduction occurs at the:
- A.anode
- B.cathode
- C.salt bridge
- D.electrolyte
2.
Faraday's first law states that mass deposited is proportional to:
- A.voltage
- B.charge passed
- C.resistance
- D.time only
3.
The electrolyte in a lead storage battery is:
- A.KOH
- B.NH4Cl paste
- C.dilute H2SO4
- D.molten NaCl
4.
Rusting of iron requires:
- A.only water
- B.only oxygen
- C.both water and oxygen
- D.neither
5.
A dry cell gives an EMF of about:
- A.1.5 V
- B.2.0 V
- C.12 V
- D.0.5 V
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State Faraday's first law of electrolysis.
7.
Calculate the charge to deposit 0.5 mol of copper ($\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$).
8.
Differentiate between a primary and a secondary cell.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Describe the working of a hydrogen–oxygen fuel cell with electrode reactions.
10.
Explain three methods of preventing the corrosion of iron.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
State Faraday's two laws of electrolysis. A current of $4\,\text{A}$ flows for $30$ minutes through $\text{CuSO}_4$ solution. Calculate the mass of copper deposited ($M = 63.5$, $n = 2$, $F = 96500$).
Answer Key
Section A — Multiple Choice Questions
- (B) cathode
- (B) charge passed
- (C) dilute H2SO4
- (C) both water and oxygen
- (A) 1.5 V
Section B — Short Answer (2 marks)
- Mass deposited is proportional to charge passed: $m = ZIt$ ($Q = It$).
- $Q = nF \times \text{moles} = 2 \times 96500 \times 0.5 = 96500\,\text{C}$.
- Primary cells are non-rechargeable (irreversible); secondary cells can be recharged (reversible).
Section C — Short Answer (3 marks)
- In KOH: anode $\text{H}_2 + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} + 2e^-$; cathode $\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-$; net product is water.
- Barrier coatings (paint/plating), galvanising with zinc, and cathodic protection using a sacrificial anode (Mg/Zn).
Section D — Long Answer (5 marks)
- First law $m = ZIt$; second law masses $\propto$ equivalent masses. $Q = 4 \times 1800 = 7200\,\text{C}$; $m = \frac{63.5 \times 7200}{2 \times 96500} = 2.37\,\text{g}$.
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